CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1
Exercise 11.1

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Answer

Steps of construction:
i. At first, we will draw a line segment $ \mathrm{AB}=7.6 \mathrm{~cm} $.

ii. Draw a ray AX such that it makes an acute angle with AB .

iii. Now, locate 13 points $ (5+8) \mathrm{A}_{1}, \mathrm{~A}_{2},
\mathrm{~A}_{3}, \ldots \ldots . \mathrm{A}_{13} $ on AX so that;
$\mathrm{AA}_{1}=\mathrm{A}_{1}\mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3}=\mathrm{A}_{3} \mathrm{~A}_{4}=\mathrm{A}_{4} \mathrm{~A}_{5}=\mathrm{A}_{5} \mathrm{~A}_{6}=\mathrm{A}_{6} \mathrm{~A}_{7}=\mathrm{A}_{7} \mathrm{~A}_{8}=\mathrm{A}_{8} \mathrm{~A}_{9}=$
$\mathrm{A}_{9} \mathrm{~A}_{10}=\mathrm{A}_{10} \mathrm{~A}_{11}=\mathrm{A}_{11} \mathrm{~A}_{12}=\mathrm{A}_{12} \mathrm{~A}_{13}$

iv. Join $ \mathrm{A}_{13} $ to B
v. Draw a line $ A_{5} C \| A_{13} B $; which intersects $ A B $ at point $ C $ and passes Through the point $ A_{5} $.
Now we have, $\mathrm{AC} : \mathrm{CB}=5: 8 $
On measuring with the scale.
$\mathrm{AC}=2.92 \mathrm{~cm}$
And
$\mathrm{CB}=4.68 \mathrm{~cm}$
Justification:
Since $ \angle \mathrm{AA}_{13} \mathrm{~B}=\angle \mathrm{AA}_{5} \mathrm{C} $
With $ A X $ is a transversal, corresponding angles are equal, Hence $ \mathrm{A}_{13} \mathrm{~B} $ and $ \mathrm{A}_{5} \mathrm{C} $ are parallel.
So by basic proportionality theorem, $ \frac{A A_{5}}{A_{5} A_{13}}=\frac{A C}{C B} $
By construction $ \frac{A A_{5}}{A_{5} A_{13}}=\frac{5}{8} $
So, $ \frac{A C}{C B}=\frac{5}{8} $
Hence C divides AB in the ratio 5:8.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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2. Construct a triangle of sides $ 4 \mathrm{~cm}, 5 \mathrm{~cm} $ and $6\mathrm{~cm}$ and then a triangle similar to it whose sides are $ \frac{2}{3} $ of the corresponding sides of the first triangle.

Answer

Steps of construction:
i. Now in order to make a triangle, draw a line segment $ \mathrm{AB}=4 \mathrm{~cm} $.

ii. Then, draw an arc at $5$ cm from point A. From point B, draw an arc at $6$ cm in such a way that it intersects the previous arc. Join the point of intersection from points A and B. This gives the required triangle ABC.

iii. Now, divide the base in the ratio $ 2: 3 $. Draw a ray $ A X $ which is at an acute angle from $ A B $. Then, plot three points on $ A X $ such that; $ \mathrm{AA}_{1}=\mathrm{A}_{1} \mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3} $. Join $ \mathrm{A}_{3} $ to B .
iv. Draw a line from point $ A_{2} $ parallel to $ A_{3} B $ and intersects $ A B $ at point $ B^{\prime} $.
v. Draw a line from point $ B^{\prime} $ that is parallel to $BC$ intersecting $AC$ at point $ C^{\prime} $.

Hence, triangle $ \mathrm{AB}^{\prime} \mathrm{C}^{\prime} $ is the required triangle.
justification:

Since the scalar factor is $ \frac{ 2 }{ 3 } $
We need to prove:
$ \frac{A B^{\prime}}{A B}=\frac{A C^{\prime}}{A C}=\frac{B^\prime C^{\prime}}{B C}=\frac{2}{3} $
By construction,
$\frac{A B^{\prime}}{A B}=\frac{A A_{2}}{A A_{3}}=\frac{2}{3} \ldots(1)$
Also, $ B^{\prime} C^{\prime} $ is parallel to $BC$.
So, both will make same angle with $AB$.
$ \therefore \angle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}=\angle \mathrm{ABC} $ (corresponding angles)
In $ \Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime} $ and $ \Delta \mathrm{ABC} $
$\angle \mathrm{A}=\angle \mathrm{A}(\text { common })$$\angle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}=\angle \mathrm{ABC}$
So, by $AA$ similarity
$\Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime} \sim \Delta \mathrm{ABC}$
As we know if two triangles are similar the ratio of their corresponding sides are also equal.
So,
$\frac{A B^{\prime}}{A B}=\frac{A C^{\prime}}{A C}=\frac{B^{\prime} C^{\prime}}{B C}$
From (1),
$\frac{A B^{\prime}}{A B}=\frac{A C^{\prime}}{A C}=\frac{B^{\prime} C^{\prime}}{B C}=\frac{2}{3}$
Hence proved.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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3. Construct a triangle with sides $ 5 \mathrm{~cm}, 6 \mathrm{~cm} $ and 7 cm and then another triangle whose sides are $ \frac{7}{5} $ of the corresponding sides of the first triangle.

Answer

1. Now in order to make a triangle, draw a line segment $ A B $ $ =5 \mathrm{~cm} $. From point A, draw an arc at 6 cm. Draw an arc at 7 cm from point $ B $ intersecting the previous arc.

2. Join the point of intersection from A and B .

Hence, this gives the required triangle ABC
3. Dividing the base, draw a ray $AX$ which is at an acute angle from AB

4. Plot seven points on $ A X $ such that:
$\mathrm{AA}_{1}=\mathrm{A}_{1} \mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3}=\mathrm{A}_{3} \mathrm{~A}_{4}=\mathrm{A}_{4} \mathrm{~A}_{5}=\mathrm{A}_{5} \mathrm{~A}_{6}=\mathrm{A}_{6} \mathrm{~A}_{7}$

5. Join $ \mathrm{A}_{5} $ to B .
6. Draw a line from point $ \mathrm{A}_{7} $ that is parallel to $ \mathrm{A}_{5} \mathrm{B} $ and joins $ \mathrm{AB}^{\prime}(\mathrm{AB} $ extended to $ \left.\mathrm{AB}^{\prime}\right) $.
7. Draw a line $ \mathrm{B}^{\prime} \mathrm{C}^{\prime} \| \mathrm{BC} $.

Hence, triangle $ \mathrm{AB}^{\prime} \mathrm{C}^{\prime} $ is the required triangle.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $ 1 \frac{1}{2} $ times the corresponding sides of the isosceles triangle.

Answer

Steps of construction:
i. Now in order to make a triangle, draw a line segment $ \mathrm{AB}=8 \mathrm{~cm} $.

ii. Draw two arcs intersecting at 4 cm distance from points A and B; on either side of AB.
Join these arcs to get perpendicular bisector CD of AB. (Since, altitude is the perpendicular bisector of base of isosceles triangle).

iii. Join points $ A $ and $ B $ to $ C $ in order to get the triangle $ A B C $.

iv. Now, draw a ray AX which is at an acute angle from point A .
As $ 1 \frac{1}{2}=\frac{3}{2} $
And 3 is greater between 3 and 2, So Plot 3 points on $AX$ such that: $ \mathrm{AA}_{1}=\mathrm{A}_{1} \mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3} $.

v. As 2 is smaller between 2 and 3 . Join $ \mathrm{A}_{2} $ to point B . Draw a line from $ \mathrm{A}_{3} $ which is parallel to $ \mathrm{A}_{2} \mathrm{~B} $ meeting the extension of $ A B $ at $ B^{\prime} $.

vi. Draw $ \mathrm{B}^{\prime} \mathrm{C}^{\prime} \| \mathrm{BC} $. Then, draw $ \mathrm{A}^{\prime} \mathrm{C}^{\prime} \| \mathrm{AC} $.

Triangle $ \mathrm{AB}^{\prime} \mathrm{C}^{\prime} $ is the required triangle.
Justification:
We need to prove,
$ \frac{A C \prime}{A C}=\frac{A B \prime}{A B}=\frac{C \prime B \prime}{C B}=\frac{3}{2} $
By construction $ \frac{A B^{\prime}}{A B}=\frac{A A_{3}}{A A_{2}}=\frac{3}{2} $ $\quad\dots.$(1)
$s C'B'\| CB$
They will maker equal angles with line $ \mathrm{AB} \angle \mathrm{ACB}=\angle \mathrm{AC}^{\prime} \mathrm{B}^{\prime} $$\quad \dots.$ (corresponding angles)
In $ \triangle \mathrm{ACB} $ and $ \triangle \mathrm{AC}^{\prime} \mathrm{B}^{\prime} $
$ \angle \mathrm{A}=\angle \mathrm{A} $ (common) $ \angle \mathrm{ACB}=\angle \mathrm{AC} \mathrm{B}^{\prime} $ (corresponding angles)
So $ \triangle \mathrm{ACB} \sim \triangle \mathrm{AC}^{\prime} \mathrm{B}^{\prime} $
As corresponding sides of similar triangles are in ratio, Hence, $ \frac{A C \prime}{A C}= $ $ \frac{A C^{\prime}}{A C}=\frac{A B \prime}{A B}=\frac{C \prime B \prime}{C B}=\frac{3}{2} $

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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5. Draw a triangle $ A B C $ with side $ B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm} $ and $ \angle A B C=60^{\circ} $. Then construct a triangle whose sides are $ \frac{3}{4} $ of the corresponding sides of the triangle ABC .

Answer

Given in $ \Delta \mathrm{ABC} $,
Length of side $ \mathrm{BC}=6 \mathrm{~cm} $.
Length of side $ \mathrm{AB}=5 \mathrm{~cm} $.
$ \angle \mathrm{ABC}=60^{\circ} $.
Steps of Construction:
1. Draw a line segment BC of length 6 cm .

2. With B as center, draw a line which makes an angle of $ 60^{\circ} $ with BC.
Construction of $ 60^{\circ} $ angle at B:
a. With B as centre and with some convenient radius draw an arc which cuts the line BC at
D.

b. With D as radius and with same radius (in step a), draw another arc which cuts the previous arc at E.

c. Join BE. The line BE makes an angle $ 60^{\circ} $ with BC.

3. Again with $ B $ as centre and with radius of 5 cm , draw an arc which intersects the line BE at point A .

4. Join AC. This is the required triangle.

5. Now, from B , draw a ray BX which makes an acute angle on the opposite side of the vertex $ A $.

6. With $ B $ as center, mark four points $ B_{1}, B_{2}, B_{3} $ and $ B_{4} $ on $ B X $ such that they are equidistant. i.e. $ \mathrm{BB}_{1}=\mathrm{B}_{1} \mathrm{~B}_{2}=\mathrm{B}_{2} \mathrm{~B}_{3}=\mathrm{B}_{3} \mathrm{~B}_{4} $.

7. Join $ B_{4} C $ and then draw a line from $ B_{3} $ parallel to $ B_{4} C $ which meets the line BC at P .

8. From $ P $, draw a line parallel to $ A C $ and meets the line $ A B $ at $ Q $. Thus $ \triangle \mathrm{BPQ} $ is the required triangle.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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6. Draw a triangle $ABC$ with side $ \mathrm{BC}=7 \mathrm{~cm}, \angle \mathrm{B}=45^{\circ}, \angle \mathrm{A}= $ $ 105^{\circ} $. Then, construct a triangle whose sides are $ \frac{4}{3} $ times the corresponding sides of $ \triangle \mathrm{ABC} $.

Answer

Steps of construction:
1. Draw a line segment $ \mathrm{BC}=7 \mathrm{~cm} $.

2. Draw $ \angle \mathrm{ABC}=45^{\circ} $ and $ \angle \mathrm{ACB}=30^{\circ} $ i.e. $ \angle \mathrm{BAC}=105^{\circ} $.

We obtain $ \triangle \mathrm{ABC} $.
3. Draw a ray $BX$ making an acute angle with BC. Mark four points $ \mathrm{B}_{1}, \mathrm{~B}_{2}, \mathrm{~B}_{3}, \mathrm{~B}_{4} $ at equal distances.

4. Through $ \mathrm{B}_{3} $ draw $ \mathrm{B}_{3} \mathrm{C} $ and through $ \mathrm{B}_{4} $ draw $ \mathrm{B}_{4} \mathrm{C}_{1} $ parallel to $ \mathrm{B}_{3} \mathrm{C} $. Then draw $ \mathrm{A}_{1} \mathrm{C}_{1} $ parallel to AC.

$ \therefore \mathrm{A}_{1} \mathrm{BC}_{1} $ is the required triangle.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm . Then construct another triangle whose sides are $ \frac{5}{3} $ times the corresponding sides of the given triangle.

Answer

Steps of construction:
1. Now in order to make a triangle, draw a line segment $ \mathrm{AB}=3 $ cm .

2. Make a right angle at point A and draw $ \mathrm{AC}=4 \mathrm{~cm} $ from this point.

3. Join points $ A $ and $ B $ to get the right triangle $ A B C $.

4. Now, Dividing the base, draw a ray $AX$ such at it forms an acute angle from $ A B $.

5. Then, plot 5 points on $ A X $ such that: $\mathrm{AG}=\mathrm{GH}=\mathrm{HI}=\mathrm{IJ}=\mathrm{JK} .$

6. Join I to point line $ A B $ and Draw a line from $K$ which is parallel to $IB$ such that it meets $AB$ at point $M$ .
7. Draw $ \mathrm{MN} \| \mathrm{CB} $.

This is the required construction, thus forming $AMN$ which have all the sides $ \frac{ 5 }{ 3 } $ times the sides of $ABC$
Triangle $AMN$ is the required triangle.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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